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Download Lesson 1 Introducing Newtons Second Law
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Transcript
Quick Starter The blocks in the diagram below are in equilibrium, g = 10ms-2 Find the friction force on the 4kg block and the tensions in the ropes. 4 kg 6 kg 3 kg Lesson Objectives: 1) Be able to recognise when a system is not in equilibrium 2) Begin to use Newton’s second law to find missing forces when an object is not in equilibrium Newton’s first law: A particle/body is in equilibrium if it is at rest or travelling at constant velocity in a straight line For each of the following situations decide if ‘Fred Lemming’ is in equilibrium or not. ‘Fred Lemming’ is standing on the floor and not moving ‘Fred Lemming’ is sitting in a car that is slowing down as some traffic lights approach ‘Fred Lemming’ is sitting at rest on a table that is sloping at 30 degrees to the horizontal ‘Fred Lemming’ is standing in a lift as it starts to move up three floors ‘Fred Lemming’ is sitting in a car that is travelling along a straight road at 30 mph ‘Fred Lemming’ is sitting on a ‘Merry-Go-Round’ as it spins round at the fun fair ‘Fred Lemming’ is falling under gravity Newton’s second law: A body that is not in equilibrium must be accelerating/decelerating The Resultant Force causing the acceleration is found using the formula: Resultant Force = Mass of the body × Acceleration OR Forces in Direction of Motion – Forces in Opposite Direction = Mass of the body × Acceleration Forces in Direction of Motion – Forces in Opposite Direction = Mass of the body × Acceleration Eg. The 2kg block below is moving to the right. Write an expression for the Resultant Force and find the acceleration. 3N 2N 2 Kg 7N Forces in Direction of Motion – Forces in Opposite Direction = Mass of the body × Acceleration Eg. The 3kg block below is moving to the left. Write an expression for the Resultant Force and find the acceleration. 4N 1N 3 kg 8N In the pack are 10 cards: Five are force diagrams Five are situation descriptors Match them up and label the forces on the diagrams! When you have finished pick a card. Find the acceleration of the object on the card and use it to find the Resistance force present in the question. Note 1: The difficulty of the questions increase with the card. Card A is the easiest and E is the hardest. Note 2: Do not start with E! A A mass of 0.5kg is at rest but falls into a pool of water 3m deep. It takes 4 seconds to reach the bottom. C D A car of mass 2000kg is travelling at 10ms-2 along a flat road when it brakes with a constant deceleration so that it can stop at some traffic lights 25m away. The braking force of the car is 3700N. 2 A rocket of mass 7,000kg starts from rest and accelerates to 30ms-1 in the time it takes to travel 50m. The driving force of the rocket is 135000N. 3 4 B A car of mass 2000kg driving along a flat road accelerates from 5ms-1 to 10ms-1 in 30 seconds. The driving force of the car is 3500N. 1 A A mass of 0.5kg is at rest but falls into a pool of water 3m deep. It takes 4 seconds to reach the bottom. v2 = u2 + 2as s = ut + ½at2 v = u + at B A car of mass 2000kg driving along a flat road accelerates from 5ms-1 to 10ms-1 in 30 seconds. The driving force of the car is 3500N. v2 = u2 + 2as s = ut + ½at2 v = u + at C A rocket of mass 7,000kg starts from rest and accelerates to 30ms-1 in the time it takes to travel 50m. The driving force of the rocket is 135000N. v2 = u2 + 2as s = ut + ½at2 v = u + at D A car of mass 2000kg is travelling at 10ms-2 along a flat road when it brakes with a constant deceleration so that it can stop at some traffic lights 25m away. The braking force of the car is 3700N. v2 = u2 + 2as s = ut + ½at2 v = u + at E A car of mass 1000kg tows a caravan of mass 750kg along a horizontal road. The engine of the car exerts a force of 2500N. The resistance force on the caravan is half of the resistance force on the car. The vehicles are travelling at 5ms-1 and accelerate to 7ms-1 in 12m. v2 = u2 + 2as s = ut + ½at2 v = u + at